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Let $V$ be a vector space over $\mathbb{C}$. Suppose $X\subset \mathbb{P} V$ is an algebraic variety, and consider its projective dual variety $X^\vee \subset \mathbb{P} V^*$. If the coordinate ring $\mathbb{C}[X]$ is arithmetically Cohen-Macaulay (aCM) is the same true for $\mathbb{C}[X^\vee]$?

An example where this is true is when $X$ is $n\times n$ matrices of rank $1$ and $X^\vee$ is $n\times n$ matrices of rank $n-1$.

I guess that it is true in general. Does someone have a reference (or a counter-example)?

Thanks!

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  • $\begingroup$ I guess I should rephrase the question: What reasonable conditions on $X$ will guarantee that $X^\vee$ is aCM? $\endgroup$ – Luke Oeding Aug 25 '14 at 16:19
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Not true. Start from a variety $Y\subset \mathbb{P}V^*$ which is not aCM -- for instance a quartic rational curve in $\mathbb{P}^3$. In most cases, in particular in the example, its dual $X:=Y^\vee$ is a hypersurface in $\mathbb{P}V$, hence is aCM. But $X^\vee=Y$ is not.

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    $\begingroup$ And is there an example with $X$ smooth? $\endgroup$ – Sasha Aug 21 '14 at 8:33
  • $\begingroup$ Good question! It would have to be quite special, since the dual variety of "most" smooth varieties is a hypersurface. $\endgroup$ – abx Aug 21 '14 at 9:02

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